Value of Keq at 25∘C for reaction : NH3+H3O+⇌NH4++H2O is 2×109. Determine pKb of NH4OH.

Ka(NH4+)= Keq1=5×10−10Now, Kb(NH3)×Ka(NH4+)=Kw=10−14.So, pKb(NH4OH)=−log10( Kb(NH3))=−log10(2×10−5)=4.7

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Ionisation of acids and bases

Value of {K}_{ {eq }} ~at~25^{} {C} for reaction : {NH}_{3}+{H

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Hard

Solving time: 5 mins

Value of $K_{eq} a t 2 5^{\circ} C$ for reaction : $NH_{3} + H_{3} O^{+} ⇌ NH_{4}^{+} + H_{2} O$ is $2 \times 1 0^{9}$ . Determine $pK_{b}$ of $NH_{4} OH$ .

4.0

4.7

3.7

Views: 5,590 students

Updated on: Aug 24, 2022

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Oliver Discussed

Value of

K_{eq} a t 2 5^{\circ} C

for reaction :

NH_{3} + H_{3} O^{+} ⇌ NH_{4}^{+} + H_{2} O

2 \times 1 0^{9}

. Determine

pK_{b}

NH_{4} OH

12 mins ago

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12 mins ago

Text solutionVerified

K_{a} (NH_{4^{+}}) = \frac{1}{K _{eq}} = 5 \times 1 0^{- 10}

Now,

K_{b} (NH_{3}) \times K_{a} (NH_{4}^{+}) = K_{w} = 1 0^{- 14}

.
So,

pK_{b} (NH_{4} OH) = - lo g_{10} (K_{b} (NH_{3})) = - lo g_{10} (2 \times 1 0^{- 5}) = 4.7

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Question Text	Value of $K_{eq} a t 2 5^{\circ} C$ for reaction : $NH_{3} + H_{3} O^{+} ⇌ NH_{4}^{+} + H_{2} O$ is $2 \times 1 0^{9}$ . Determine $pK_{b}$ of $NH_{4} OH$ .
Updated On	Aug 24, 2022
Topic	Equilibrium
Subject	Chemistry
Class	Class 11
Answer Type	Text solution:1 Video solution: 7
Upvotes	876
Avg. Video Duration	5 min

Value of $K_{eq} a t 2 5^{\circ} C$ for reaction : $NH_{3} + H_{3} O^{+} ⇌ NH_{4}^{+} + H_{2} O$ is $2 \times 1 0^{9}$ . Determine $pK_{b}$ of $NH_{4} OH$ .

Text solutionVerified

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Value of Keq ​ at 25∘C for reaction : NH3​+H3​O+⇌NH4​++H2​O is 2×109. Determine pKb​ of NH4​OH.

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Value of $K_{eq} a t 2 5^{\circ} C$ for reaction : $NH_{3} + H_{3} O^{+} ⇌ NH_{4}^{+} + H_{2} O$ is $2 \times 1 0^{9}$ . Determine $pK_{b}$ of $NH_{4} OH$ .